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Paha777 [63]
1 year ago
8

Which of following is not malicious ? Worm Trogan Horse Driver Virus

Engineering
1 answer:
Vesna [10]1 year ago
7 0

Driver is not considered as malicious .  Malicious   includes computer viruses, worms, trojan horses, rootkits, spyware, adware etc.

<h3>Malicious software :</h3>

Malware is unwanted software that an unauthorized person tries to run on your computer. These are known as security threats. These include computer viruses, worms, Trojan horses, rootkits, spyware, adware, and more. Malware types include computer viruses, worms, Trojan horses, ransomware, and spyware. These malicious programs steal, encrypt and delete sensitive data. Modify or hijack core computing functionality and monitor end-user computing activity

<h3>Are Trojan Horses Malware?</h3>

A Trojan horse (Trojan horse) is a type of malware that disguises itself as legitimate code or software. Once inside the network, the attacker can take any action a legitimate user can take, including: exporting files, modifying data, deleting files or otherwise altering the contents of the device

Learn more about Malicious software :

brainly.com/question/29662363

#SPJ4

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The flatbed truck carries a large section of circular pipe secured only by the two fixed blocks A and B of height h. The truck i
zalisa [80]

Answer:

The maximum speed for which the pipe will be restrained is 18.05 m/s.

Explanation:

We note that the

Securing blocks have height, h = 0.1 m

The radius of travel, rho = 60 m

The radius of the circular pipe, R = 0.8 m

We note that the weight of the pipe is acting at the centroid of the pipe

Therefore, for the pipe to slip, it has to climb the wedge h

Taking moments about h, we have  

mg × R sin α = ma × R cos α  

a/g  =  Tan α

But a = \frac{V^{2} }{rho}

Therefore  \frac{V^{2} }{rho} = g tanα

Since height of the block, h = 0.1 m therefore,

R cos α = R - h

That is 0.8 cos α = 0.8 - 0.1 = 0.7

Therefore α = cos⁻¹ (0.7/0.8) = 28.96 °

From which V² = rho × g× tanα = 60 × 9.81 × tan 28.96

= 325.66 m²/s²

∴ V = √(325.66 m²/s²)  = 18.05 m/s

Maximum speed = 18.05 m/s.

6 0
3 years ago
Read 2 more answers
Oil, with density of 900 kg/m3 and kinematic viscosity of 0.00001 m2/s, flows at 0.2 m3/s through 500 m of 200-mm-diameter cast-
Vsevolod [243]
Just trying to personally understand this is making me shed a few tears. Good luck with finding someone to answer your question!
5 0
3 years ago
Write a matrix, that is a lower triangular matrix.
shepuryov [24]

Answer:

\left[\begin{array}{ccc}10&0&0\\14&25&0\\57&18&39\end{array}\right]

Explanation:

A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of  the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.

If there is any question, just let me know.

6 0
4 years ago
The kinetic energy correction factor depends on the (shape — volume - mass) of the cross section Of the pipe and the (velocity —
butalik [34]

Answer:

The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.

Explanation:

The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform.  In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.

3 0
3 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
3 years ago
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